How would I find the residue of $\text{sech}$ and $\coth$ at their poles? I thought I had understood this, but I'm now lost when trig. functions are introduced and I don't know how to continue. I attempted to apply the $\lim_{z \to a} (z-a)f(z)$ on it, but that didn't take me far.
 A: The residues of $\coth$ are simple, since
$$\coth z = \frac{\cosh z}{\sinh z} = \frac{\sinh' z}{\sinh z},$$
and the residue of $\dfrac{f'(z)}{f(z)}$ is always the multiplicity of the zero of $f$ (poles are counted as zeros of negative multiplicity). Since $\sinh$ has only simple zeros, all residues of $\coth$ are $1$. The poles of $\coth$ are the zeros of $\sinh$, that is $k\pi i$ for $k \in \mathbb{Z}$.
For $\operatorname{sech}$, the matter is similar, but not quite as easy. If $f$ has a simple zero in $z_0$, the residue of $\dfrac1f$ in $z_0$ is $\dfrac{1}{f'(z_0)}$. Since $\cosh' = \sinh$, the zeros of $\operatorname{sech}$ are then the reciprocal values of $\sinh$ in the zeros of $\cosh$. Since $\cosh^2 z - \sinh^2 z \equiv 1$, $\sinh z_0 = \pm i$ when $\cosh z_0 = 0$, and it remains to find the sign. The poles of $\operatorname{sech}$ are the zeros of $\cosh$, that is $(k+\frac12)\pi i$ for $k\in\mathbb{Z}$. We have $\sinh \frac{\pi i}{2} = i\sin \frac{\pi}{2} = i$, so the residue of $\operatorname{sech}$ in $\frac{\pi i}{2}$ is $\frac{1}{i}$, and since the sign of $\sinh z_k$ alternates, we have
$$\operatorname{Res}\left(\operatorname{sech};z_k \right) = \frac{(-1)^k}{i},$$
where $z_k = (k+\frac12)\pi i$.
